Multiplying Mixed Numbers

Date: 06/09/98 at 21:58:08
From: Doctor Peterson
Subject: Re: mult. fractions
Hi, Devon. Fractions can be a lot of fun once you've made friends with
them, but they can be a little scary when you first meet them. These
two look pretty ugly, so we'd better straighten them up before we try
to work with them. In order to give you a chance to practice with your
own problem, I won't do exactly your problem, but one like it:
2 3/7 x 3 2/3
To make them look neater, we first change the mixed numbers into
improper fractions. You probably remember how to do that. You want to
turn 2 3/7 into some number of sevenths, so you remember that 2 is
just 14 sevenths, and add them:
3 14 3 17
2 --- = --- + --- = ---
7 7 7 7
The same way,
2 9 2 11
3 --- = --- + --- = ---
3 3 3 3
Note: When you're used to it, you can try my way, which is to start at
the denominator, multiply by the number clockwise from it (the whole
number) and add the next number clockwise (the numerator). This way,
we get:
3 x 3 + 2 = 11
by following the arrow:
+-------->
| + 2 -> 11
| 3 --- --
| x 3 -> 3
+-----
But that's just a trick to avoid extra writing.
Okay, so regardless of how you change the mixed numbers into
improper fractions, our problem is really:
20 11
--- x ---
7 3
Now how do you multiply fractions? The "rule" says you multiply the
numerators to get the numerator of the product, and you multiply the
denominators to get the denominator of the product:
a c a x c
- x - = -----
b d b x d
So in this case, we multiply 20 x 11 and 7 x 3:
20 11 20 x 11 220
--- x --- = ------- = ---
7 3 7 x 3 21
Often, you'll find something to cancel out to put the result in lowest
terms; I recommend doing that before you actually multiply anything,
to save work. In this case, we don't find anything to cancel, so we'll
try the cancelling trick in the next example.
Now I never like to do something just because of a rule, so I'll let
you in on the secret behind that rule. Let's take a simpler example,
2 3
- x -
3 5
I can draw a picture of 2/3 by dividing a rectangle into three (the
denominator) slices and coloring in two (the numerator) of them:
+--------------+
|//////////////|
|//////////////|
+--------------+
|//////////////|
|//////////////|
+--------------+
| |
| |
+--------------+
Now I want 3/5 of what I colored in. To do that, I can cut the
whole thing into five slices and take three of the pieces that are
colored in:
+--+--+--+--+--+
|XX|XX|XX|//|//|
|XX|XX|XX|//|//|
+--+--+--+--+--+
|XX|XX|XX|//|//|
|XX|XX|XX|//|//|
+--+--+--+--+--+
| | | | | |
| | | | | |
+--+--+--+--+--+
You can see I've cut the original rectangle into 3 x 5 = 15 pieces
(that's the denominator of the result, which is the product of the
denominators), and I've picked 2 x 3 = 6 of them (that's the
numerator, which is the product of the two numerators). So 3/5 of 2/3
is 6/15. I can rearrange the six pieces to simplify the fraction and
show that it's really 2/5:
+--+--+--+--+--+
|XX|XX| | | |
|XX|XX| | | |
+--+--+--+--+--+
|XX|XX| | | |
|XX|XX| | | |
+--+--+--+--+--+
|XX|XX| | | |
|XX|XX| | | |
+--+--+--+--+--+
So when you follow the rule, you're really just cutting and choosing,
then cutting and choosing again.
To check our example, we can use the rule. In this case, we multiply
2 x 3 for the numerator, and 3 x 5 for the denominator:
2 3 2 x 3 6
- x - = ----- = --
3 5 3 x 5 15
Now we can reduce the fraction to lowest terms, dividing the numerator
and denominator by 3 to get 2/5. That's not really part of
multiplying, but it makes the result easier to work with.
Something that often saves work, though, is to simplify your fraction
before you actually do the multiplication. That's because simplifying
means looking for factors you can cancel, and the factors are easier
to see before you multiply them. In this example, you can see the
two 3's sitting there, and cancel them without ever having to multiply
at all:
/
2 3 2 x 3 2
- x - = ----- = -
3 5 3 x 5 5
/
See if that helps you work out your problem, and let us know if you
need more.
-Doctor Peterson, The Math Forum
Check out our web site! http://mathforum.org/dr.math/